Let the matrix A =−1 0 00 0 01 0 1.(a) (5 pts) Why is A diagonalizeable?(b) (12 pts) Diagonalize A and find the invertible matrix P by which you diagonalized A.2
Question
Let the matrix A =−1 0 00 0 01 0 1.(a) (5 pts) Why is A diagonalizeable?(b) (12 pts) Diagonalize A and find the invertible matrix P by which you diagonalized A.2
Solution
(a) A matrix is diagonalizable if it has enough eigenvectors to form a basis for the space. In this case, the matrix A is already a diagonal matrix. Diagonal matrices are always diagonalizable because the eigenvectors can be the standard basis vectors.
(b) To diagonalize a matrix, we need to find the eigenvalues and eigenvectors.
For the given matrix A, the eigenvalues can be found by solving the characteristic equation, det(A - λI) = 0, where I is the identity matrix and λ are the eigenvalues.
The matrix A - λI is:
[-1-λ 0 0] [0 -λ 0] [1 0 1-λ]
Setting the determinant equal to zero gives the equation (-1-λ)(-λ) (1-λ). Solving this equation gives the eigenvalues λ1 = -1, λ2 = 0, λ3 = 1.
The corresponding eigenvectors are found by plugging each eigenvalue back into the equation (A - λI)v = 0 and solving for v.
For λ1 = -1, we get the eigenvector v1 = [1, 0, 0]. For λ2 = 0, we get the eigenvector v2 = [0, 1, 0]. For λ3 = 1, we get the eigenvector v3 = [0, 0, 1].
The matrix P that diagonalizes A is formed by placing the eigenvectors v1, v2, v3 as columns:
P = [1 0 0] [0 1 0] [0 0 1]
So, the diagonalized form of A is a diagonal matrix with the eigenvalues on the diagonal, and the matrix P is the identity matrix.
Similar Questions
diagonalizing a matrix;
The matrix A𝐴 has eigenvalues −5−5, −1−1, and 22 with corresponding eigenvectors ⎡⎣⎢07−6⎤⎦⎥[07−6], ⎡⎣⎢−13−1⎤⎦⎥[−13−1] and ⎡⎣⎢−764⎤⎦⎥[−764]. Which of the following is a valid diagonalisation? Select all that apply.This matrix cannot be diagonalisedA=V−1DV𝐴=𝑉−1𝐷𝑉 with V=𝑉= ⎡⎣⎢07−6−13−1−764⎤⎦⎥[0−1−7736−6−14] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥[−5000−10002]A=V−1DV𝐴=𝑉−1𝐷𝑉 with V=𝑉= ⎡⎣⎢0−1−7736−6−14⎤⎦⎥[07−6−13−1−764] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥[−5000−10002]A=VDV−1𝐴=𝑉𝐷𝑉−1 with V=𝑉= ⎡⎣⎢0−1−7736−6−14⎤⎦⎥[07−6−13−1−764] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥[−5000−10002]A=VDV−1𝐴=𝑉𝐷𝑉−1 with V=𝑉= ⎡⎣⎢07−6−13−1−764⎤⎦⎥[0−1−7736−6−14] and D=𝐷= ⎡⎣⎢−5000−10002⎤⎦⎥
Suppose that a 2×2 matrix A has an eigenvalue 2 with corresponding eigenvector [1 −1] and an eigenvalue −2 with corresponding eigenvector [3 −2]. Find an invertible matrix P and a diagonal matrix D so that A=PDP−1. Enter your answer as an equation of the form A=PDP−1. You must enter a number in every answer blank for the answer evaluator to work properly.
For the matrix A, find (if possible) a nonsingular matrix P such that P−1AP is diagonal. (If not possible, enter IMPOSSIBLE.)A = 2 −2 9 0 3 −20 −1 2
Time left 0:11:05Question 2Tries remaining: 2Marked out of 1.00Flag questionQuestion textDiagonalize the matrix [−1−425][−12−45].That is, find a diagonal matrix D𝐷 and an invertible matrix P𝑃 such that A=P⋅D⋅P−1
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.